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 We have already seen that as spheres cannot fill entire space → the packing
fraction (PF) < 1 (for all crystals)
 This implies there are voids between the atoms. Lower the PF, larger the volume
occupied by voids.
 These voids have complicated shapes; but we are mostly interested in the largest
sphere which can fit into these voids→ hence the plane faced polyhedron version
of the voids is only (typically) considered.
 The size and distribution of voids in materials play a role in determining aspects
of material behaviour → e.g. solubility of interstitials and their diffusivity
 The position of the voids of a particular type will be consistent with the
symmetry of the crystal
 In the close packed crystals (FCC, HCP) there are two types of voids →
tetrahedral and octahedral voids (identical in both the structures as the voids are
formed between two layers of atoms)
 The octahedral void has a coordination number 6 (should not be confused with 8 coordination!)
 In the ‘BCC crystal’ the voids do NOT have the shape of the regular tetrahedron
or the regular octahedron (in fact the octahedral void is a ‘linear void’!!)
Voids
SC
 The simple cubic crystal (monoatomic decoration of the simple cubic lattice) has large void in the centre of the unit
cell with a coordination number of 8.
 The actual space of the void in very complicated (right hand figure below) and the polyhedron version of the void is
the cube (as cube is the coordination polyhedron around a atom sitting in the void)
( 3 1) 0.732xr
r
= − =
Actual shape of the void (space)!Polyhedral model (Cube)True Unit Cell of SC crystal
Video: void in SC crystal
Video: void in SC crystal
FCC
 Actual shape of the void is as shown below.
Actual shape of void
Position of some of the atoms w.r.t to the void
FCC
 The complicated void shown before is broken down in the polyhedral representation into two shapes the octahedron
and the tetrahedron- which together fill space.
Octahedra and tetrahedra in an unit cell
Quarter of a octahedron belong to an unit cell
4-Quarters forming a full octahedron
4-tetrahedra in view
Central octahedron in view
Video of the construction
shown in this slide
Video of the construction
shown in this slide
VOIDS
Tetrahedral Octahedral
FCC = CCP
Note: Atoms are coloured differently but are the same
cellntetrahedro VV
24
1
= celloctahedron VV
6
1
=
¼ way along body diagonal
{¼, ¼, ¼}, {¾, ¾, ¾}
+ face centering translations
At body centre
{½, ½, ½}
+ face centering translations
OVTV
rvoid / ratom = 0.225
rVoid / ratom = 0.414
Video: voids CCP
Video: voids CCP
Video: atoms forming the voids
Video: atoms forming the voids
FCC- OCTAHEDRAL
{½, ½, ½} + {½, ½, 0} = {1, 1, ½} ≡ {0, 0, ½}
Face centering translation
Note: Atoms are coloured differently but are the same
Equivalent site for an
octahedral void
Site for octahedral void
Once we know the position of a void then we can use
the symmetry operations of the crystal to locate the
other voids. This includes lattice translations
FCC voids Position Voids / cell Voids / atom
Tetrahedral
¼ way from each vertex of the cube
along body diagonal <111>
→ ((¼, ¼, ¼))
8 2
Octahedral
• Body centre: 1 → (½, ½, ½)
• Edge centre: (12/4 = 3) → (½, 0,
0)
4 1
Size of the largest atom which can fit into the tetrahedral void of FCC
CV = r + x
Radius of the
new atom
e
xre +=
4
6
225.0~1
2
3
2 







−=⇒=
r
x
re
Size of the largest atom which can fit into the Octahedral void of FCC
2r + 2x = a
ra 42 =
( ) 414.0~12 −=
r
x
Thus, the octahedral void is the bigger
one and interstitial atoms (which are
usually bigger than the voids) would
prefer to sit here
VOIDS
TETRAHEDRAL OCTAHEDRAL
HCP
 These voids are identical to the ones found in FCC (for ideal c/a ratio)
 When the c/a ratio is non-ideal then the octahedra and tetrahedra are distorted (non-regular)
Important Note: often in these discussions an ideal c/a ratio will be assumed (without stating the same explicitly)
Note: Atoms are coloured differently but are the same
Coordinates: (⅓ ⅔,¼), (⅓,⅔,¾)),,(),,,(),,0,0(),,0,0(: 8
7
3
1
3
2
8
1
3
1
3
2
8
5
8
3sCoordinate
Octahedral voids occur in 1 orientation, tetrahedral voids occur in 2 orientations
The other orientation of the tetrahedral void
Note: Atoms are coloured differently but are the same
Note: Atoms are coloured differently but are the same
Further views
Note: Atoms are coloured differently but are the same
Octahedral voids
Tetrahedral void
Further views
HCP voids Position
Voids /
cell
Voids / atom
Tetrahedral
(0,0,3/8), (0,0,5/8), (⅔, ⅓,1/8),
(⅔,⅓,7/8)
4 2
Octahedral • (⅓ ⅔,¼), (⅓,⅔,¾) 2 1
Voids/atom: FCC ≡ HCP
→ as we can go from FCC to HCP (and vice-
versa) by a twist of 60° around a central atom of
two void layers (with axis ⊥ to figure) Central atom
Check below
Atoms in HCP crystal: (0,0,0), (⅔, ⅓,½)
A
A
B
Further views
Various sections along the c-axis
of the unit cell
Octahedral void
Tetrahedral void
Further views with some models
Visualizing these voids can sometimes be difficult especially in the HCP
crystal. ‘How the tetrahedral and octahedral void fill space?’ is shown in the
accompanying video
Video: Polyhedral voids filling space
Video: Polyhedral voids filling space
 There are NO voids in a ‘BCC crystal’ which have the shape of a regular
polyhedron (one of the 5 Platonic solids)
 The voids in BCC crystal are: distorted ‘octahedral’ and distorted tetrahedral
 However, the ‘distortions’ are ‘pretty regular’ as we shall see
 The distorted octahedral void is in a sense a ‘linear void’
→ an sphere of correct size sitting in the void touches only two of the six atoms surrounding it
 Carbon prefers to sit in this smaller ‘octahedral void’ for reasons which we shall
see soon
Voids in BCC crystal
VOIDS
Distorted TETRAHEDRAL Distorted OCTAHEDRAL**
BCC
a
a√3/2
a a√3/2
rvoid / ratom = 0.29 rVoid / ratom = 0.155
Note: Atoms are coloured differently but are the same ** Actually an atom of correct size touches only
the top and bottom atoms
Coordinates of the void:
{½, 0, ¼} (four on each face) Coordinates of the void:
{½, ½, 0} (+ BCC translations: {0, 0, ½})
Illustration on one face only
BCC voids Position
Voids /
cell
Voids
/ atom
Distorted
Tetrahedral
• Four on each face: [(4/2) × 6 = 12] → (0, ½, ¼) 12 6
Distorted
Octahedral
• Face centre: (6/2 = 3) → (½, ½, 0)
• Edge centre: (12/4 = 3) → (½, 0, 0)
6 3
{0, 0, ½})
Illustration on one face only
OVTV
From the right angled triange OCM:
416
22
aa
OC +=
5
4
a r x= = +
For a BCC structure: 3 4a r= (
3
4r
a = )
xr
r
+=
3
4
4
5
⇒ 29.01
3
5
=







−=
r
x
a
a√3/2
BCC: Distorted Tetrahedral Void
alculation of the size of the distorted tetrahedral void
2
a
xrOB =+=
32
4r
xr =+ raBCC 43: =
1547.01
3
32
=







−=
r
x
Distorted Octahedral Void
a√3/2
a
a
a
OB 5.0
2
== a
a
OA 707.
2
2
==
As the distance OA > OB the atom in the
void touches only the atom at B (body
centre).
⇒ void is actually a ‘linear’ void*
This implies:
alculation of the size of the distorted octahedral void
* Point regarding ‘Linear Void’
 Because of this aspect the OV along the 3 axes can be
differentiated into OVx, OVy & OVz
 Similarly the TV along x,y,z can be differentiated
 Fe carbon alloys are important materials and hence we consider them next
 The octahedral void in FCC is the larger one and less distortion occurs when
carbon sits there → this aspect contributes to a higher solubility of C in γ-Fe
 The distorted octahedral void in BCC is the smaller one → but (surprisingly)
carbon sits there in preference to the distorted tetrahedral void (the bigger one) -
(we shall see the reason shortly)
 Due to small size of the voids in BCC the distortion caused is more and the
solubility of C in α-Fe is small
 this is rather surprising at a first glance as BCC is the more open structure
 but we have already seen that the number of voids in BCC is more than that in
FCC → i.e. BCC has more number of smaller voids
See next slide for figures

A292.1=Fe
FCCr

A534.0)( =octxFe
FCC

A77.0=C
r
C
N
Void (Oct)
FeFCC
H

A258.1=Fe
BCCr 
A364.0).( =tetdxFe
BCC

A195.0).( =octdxFe
BCC
FCC
BCC
FeBCC
Relative sizes of voids w.r.t to atoms
( . )
0.155
Fe
BCC
Fe
BCC
x d oct
r
=
( . )
0.29
Fe
BCC
Fe
BCC
x d tet
r
=
Relative size of voids, interstitials and Fe atom
Note the difference in size of the atoms
Spend some time over this slide
Size of the OV
Size of Carbon atom
Size of Fe atom
CCP crystal
Size of Fe atom
BCC crystal
Size of the OV
Size of the TV
0.71 AN
r =

0.46 AH
r =

Void (Tet)
 We had mentioned that the octahedral void in BCC is a linear one
(interstitial atom actually touches only two out of the 6 atoms surrounding it)
 In the next slide we make a approximate calculation to see till what size will it
continue to touch only two Fe atoms
(these are ‘ideal’ simplified geometrical calculations and in reality other complications will have to be considered)

A258.1=Fe
BCCr
2
2
A
a
OA r x= + =
2 6
3
A
r
r x+ =
raBCC 43: =
2 6
1 0.6329
3
Ax
r
 
= − = ÷ ÷
 
Ignoring the atom sitting at B and assuming the interstitial atom touches the atom at A
0.796AAOX x= =

0.195ABOY x= =
 
A364.0).( =tetdxFe
BCC
 This implies for x/r ratios between 0.15 and 0.63 the interstitial atom has to push
only two atoms
 (xcarbon/rFe)BCC ~ 0.6
 This explains why Carbon preferentially sits in the apparently smaller octahedral
void in BCC
rvoid / ratom
SC BCC FCC DC
Octahedral
(CN = 6)
0.155
(distorted)
0.414 -
Tetrahedral
(CN = 4)
0.29
(distorted)
0.225
1
(½,½,½) & (¼, ¼, ¼)
Cubic
(CN = 8)
0.732
Summary of void sizes

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Voids in crystals

  • 1.  We have already seen that as spheres cannot fill entire space → the packing fraction (PF) < 1 (for all crystals)  This implies there are voids between the atoms. Lower the PF, larger the volume occupied by voids.  These voids have complicated shapes; but we are mostly interested in the largest sphere which can fit into these voids→ hence the plane faced polyhedron version of the voids is only (typically) considered.  The size and distribution of voids in materials play a role in determining aspects of material behaviour → e.g. solubility of interstitials and their diffusivity  The position of the voids of a particular type will be consistent with the symmetry of the crystal  In the close packed crystals (FCC, HCP) there are two types of voids → tetrahedral and octahedral voids (identical in both the structures as the voids are formed between two layers of atoms)  The octahedral void has a coordination number 6 (should not be confused with 8 coordination!)  In the ‘BCC crystal’ the voids do NOT have the shape of the regular tetrahedron or the regular octahedron (in fact the octahedral void is a ‘linear void’!!) Voids
  • 2. SC  The simple cubic crystal (monoatomic decoration of the simple cubic lattice) has large void in the centre of the unit cell with a coordination number of 8.  The actual space of the void in very complicated (right hand figure below) and the polyhedron version of the void is the cube (as cube is the coordination polyhedron around a atom sitting in the void) ( 3 1) 0.732xr r = − = Actual shape of the void (space)!Polyhedral model (Cube)True Unit Cell of SC crystal Video: void in SC crystal Video: void in SC crystal
  • 3. FCC  Actual shape of the void is as shown below. Actual shape of void Position of some of the atoms w.r.t to the void
  • 4. FCC  The complicated void shown before is broken down in the polyhedral representation into two shapes the octahedron and the tetrahedron- which together fill space. Octahedra and tetrahedra in an unit cell Quarter of a octahedron belong to an unit cell 4-Quarters forming a full octahedron 4-tetrahedra in view Central octahedron in view Video of the construction shown in this slide Video of the construction shown in this slide
  • 5. VOIDS Tetrahedral Octahedral FCC = CCP Note: Atoms are coloured differently but are the same cellntetrahedro VV 24 1 = celloctahedron VV 6 1 = ¼ way along body diagonal {¼, ¼, ¼}, {¾, ¾, ¾} + face centering translations At body centre {½, ½, ½} + face centering translations OVTV rvoid / ratom = 0.225 rVoid / ratom = 0.414 Video: voids CCP Video: voids CCP Video: atoms forming the voids Video: atoms forming the voids
  • 6. FCC- OCTAHEDRAL {½, ½, ½} + {½, ½, 0} = {1, 1, ½} ≡ {0, 0, ½} Face centering translation Note: Atoms are coloured differently but are the same Equivalent site for an octahedral void Site for octahedral void Once we know the position of a void then we can use the symmetry operations of the crystal to locate the other voids. This includes lattice translations
  • 7. FCC voids Position Voids / cell Voids / atom Tetrahedral ¼ way from each vertex of the cube along body diagonal <111> → ((¼, ¼, ¼)) 8 2 Octahedral • Body centre: 1 → (½, ½, ½) • Edge centre: (12/4 = 3) → (½, 0, 0) 4 1
  • 8. Size of the largest atom which can fit into the tetrahedral void of FCC CV = r + x Radius of the new atom e xre += 4 6 225.0~1 2 3 2         −=⇒= r x re Size of the largest atom which can fit into the Octahedral void of FCC 2r + 2x = a ra 42 = ( ) 414.0~12 −= r x Thus, the octahedral void is the bigger one and interstitial atoms (which are usually bigger than the voids) would prefer to sit here
  • 9. VOIDS TETRAHEDRAL OCTAHEDRAL HCP  These voids are identical to the ones found in FCC (for ideal c/a ratio)  When the c/a ratio is non-ideal then the octahedra and tetrahedra are distorted (non-regular) Important Note: often in these discussions an ideal c/a ratio will be assumed (without stating the same explicitly) Note: Atoms are coloured differently but are the same Coordinates: (⅓ ⅔,¼), (⅓,⅔,¾)),,(),,,(),,0,0(),,0,0(: 8 7 3 1 3 2 8 1 3 1 3 2 8 5 8 3sCoordinate
  • 10. Octahedral voids occur in 1 orientation, tetrahedral voids occur in 2 orientations The other orientation of the tetrahedral void Note: Atoms are coloured differently but are the same
  • 11. Note: Atoms are coloured differently but are the same Further views
  • 12. Note: Atoms are coloured differently but are the same Octahedral voids Tetrahedral void Further views
  • 13. HCP voids Position Voids / cell Voids / atom Tetrahedral (0,0,3/8), (0,0,5/8), (⅔, ⅓,1/8), (⅔,⅓,7/8) 4 2 Octahedral • (⅓ ⅔,¼), (⅓,⅔,¾) 2 1 Voids/atom: FCC ≡ HCP → as we can go from FCC to HCP (and vice- versa) by a twist of 60° around a central atom of two void layers (with axis ⊥ to figure) Central atom Check below Atoms in HCP crystal: (0,0,0), (⅔, ⅓,½)
  • 14. A A B Further views Various sections along the c-axis of the unit cell Octahedral void Tetrahedral void
  • 15. Further views with some models Visualizing these voids can sometimes be difficult especially in the HCP crystal. ‘How the tetrahedral and octahedral void fill space?’ is shown in the accompanying video Video: Polyhedral voids filling space Video: Polyhedral voids filling space
  • 16.  There are NO voids in a ‘BCC crystal’ which have the shape of a regular polyhedron (one of the 5 Platonic solids)  The voids in BCC crystal are: distorted ‘octahedral’ and distorted tetrahedral  However, the ‘distortions’ are ‘pretty regular’ as we shall see  The distorted octahedral void is in a sense a ‘linear void’ → an sphere of correct size sitting in the void touches only two of the six atoms surrounding it  Carbon prefers to sit in this smaller ‘octahedral void’ for reasons which we shall see soon Voids in BCC crystal
  • 17. VOIDS Distorted TETRAHEDRAL Distorted OCTAHEDRAL** BCC a a√3/2 a a√3/2 rvoid / ratom = 0.29 rVoid / ratom = 0.155 Note: Atoms are coloured differently but are the same ** Actually an atom of correct size touches only the top and bottom atoms Coordinates of the void: {½, 0, ¼} (four on each face) Coordinates of the void: {½, ½, 0} (+ BCC translations: {0, 0, ½}) Illustration on one face only
  • 18. BCC voids Position Voids / cell Voids / atom Distorted Tetrahedral • Four on each face: [(4/2) × 6 = 12] → (0, ½, ¼) 12 6 Distorted Octahedral • Face centre: (6/2 = 3) → (½, ½, 0) • Edge centre: (12/4 = 3) → (½, 0, 0) 6 3 {0, 0, ½}) Illustration on one face only OVTV
  • 19. From the right angled triange OCM: 416 22 aa OC += 5 4 a r x= = + For a BCC structure: 3 4a r= ( 3 4r a = ) xr r += 3 4 4 5 ⇒ 29.01 3 5 =        −= r x a a√3/2 BCC: Distorted Tetrahedral Void alculation of the size of the distorted tetrahedral void
  • 20. 2 a xrOB =+= 32 4r xr =+ raBCC 43: = 1547.01 3 32 =        −= r x Distorted Octahedral Void a√3/2 a a a OB 5.0 2 == a a OA 707. 2 2 == As the distance OA > OB the atom in the void touches only the atom at B (body centre). ⇒ void is actually a ‘linear’ void* This implies: alculation of the size of the distorted octahedral void * Point regarding ‘Linear Void’  Because of this aspect the OV along the 3 axes can be differentiated into OVx, OVy & OVz  Similarly the TV along x,y,z can be differentiated
  • 21.  Fe carbon alloys are important materials and hence we consider them next  The octahedral void in FCC is the larger one and less distortion occurs when carbon sits there → this aspect contributes to a higher solubility of C in γ-Fe  The distorted octahedral void in BCC is the smaller one → but (surprisingly) carbon sits there in preference to the distorted tetrahedral void (the bigger one) - (we shall see the reason shortly)  Due to small size of the voids in BCC the distortion caused is more and the solubility of C in α-Fe is small  this is rather surprising at a first glance as BCC is the more open structure  but we have already seen that the number of voids in BCC is more than that in FCC → i.e. BCC has more number of smaller voids See next slide for figures
  • 22.  A292.1=Fe FCCr  A534.0)( =octxFe FCC  A77.0=C r C N Void (Oct) FeFCC H  A258.1=Fe BCCr  A364.0).( =tetdxFe BCC  A195.0).( =octdxFe BCC FCC BCC FeBCC Relative sizes of voids w.r.t to atoms ( . ) 0.155 Fe BCC Fe BCC x d oct r = ( . ) 0.29 Fe BCC Fe BCC x d tet r = Relative size of voids, interstitials and Fe atom Note the difference in size of the atoms Spend some time over this slide Size of the OV Size of Carbon atom Size of Fe atom CCP crystal Size of Fe atom BCC crystal Size of the OV Size of the TV 0.71 AN r =  0.46 AH r =  Void (Tet)
  • 23.  We had mentioned that the octahedral void in BCC is a linear one (interstitial atom actually touches only two out of the 6 atoms surrounding it)  In the next slide we make a approximate calculation to see till what size will it continue to touch only two Fe atoms (these are ‘ideal’ simplified geometrical calculations and in reality other complications will have to be considered)
  • 24.  A258.1=Fe BCCr 2 2 A a OA r x= + = 2 6 3 A r r x+ = raBCC 43: = 2 6 1 0.6329 3 Ax r   = − = ÷ ÷   Ignoring the atom sitting at B and assuming the interstitial atom touches the atom at A 0.796AAOX x= =  0.195ABOY x= =   A364.0).( =tetdxFe BCC
  • 25.  This implies for x/r ratios between 0.15 and 0.63 the interstitial atom has to push only two atoms  (xcarbon/rFe)BCC ~ 0.6  This explains why Carbon preferentially sits in the apparently smaller octahedral void in BCC
  • 26. rvoid / ratom SC BCC FCC DC Octahedral (CN = 6) 0.155 (distorted) 0.414 - Tetrahedral (CN = 4) 0.29 (distorted) 0.225 1 (½,½,½) & (¼, ¼, ¼) Cubic (CN = 8) 0.732 Summary of void sizes